Your friend used long division and concluded that the decimal expansion for 16 is 6. What is the correct decimal expansion?
1 answer:
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There it is, I hope it gets to be helpful
Answer:
1.38x10⁻⁴ M
Explanation:
First, we <u>calculate the pOH of the solution</u>, using the <em>given pH</em>:
- pOH = 14 - pH
- pOH = 14 - 10.44
- pOH = 3.56
Now we <u>calculate the concentration of OH⁻</u>, using<em> the pOH</em>:
- pOH = -log[OH⁻]
- [OH⁻] =
= 2.75x10⁻⁴ M
Following the reaction Ca(OH)₂ → Ca⁺² + 2OH⁻, we divide [OH⁻] by 2 to calculate [Ca(OH)₂]:
- [Ca(OH)₂] = 2.75x10⁻⁴ M / 2
- [Ca(OH)₂] = 1.38x10⁻⁴ M
Taking into account the reaction stoichiometry, 1.729 grams of NaCl is formed.
<h3>Reaction stoichiometry</h3>In first place, the balanced reaction is:
CuCl₂ + 2 NaNO₃ → Cu(NO₃)₂ + 2 NaCl
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- CuCl₂: 1 mole
- NaNO₃: 2 moles
- Cu(NO₃)₂ : 1 mole
- NaCl: 2 moles
The molar mass of the compounds is:
- CuCl₂: 134.44 g/mole
- NaNO₃: 85 g/mole
- Cu(NO₃)₂ : 187.54 g/mole
- NaCl: 58.45 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- CuCl₂: 1 mole ×134.44 g/mole= 134.44 grams
- NaNO₃: 2 moles ×85 g/mole= 170 grams
- Cu(NO₃)₂ : 1 mole ×187.54 g/mole= 187.54 grams
- NaCl: 2 moles ×58.45 g/mole= 116.9 grams
The following rule of three can be applied: If by stoichiometry of the reaction 134.44 grams of CuCl₂ form 116.9 grams of NaCl, 2 grams of CuCl₂ form how much mass of NaCl?
<u><em>mass of NaCl= 1.739 grams</em></u>
Finally, 1.729 grams of NaCl is formed.
Learn more about the reaction stoichiometry: